chem calc (1 Viewer)

fullonoob · Mar 21, 2010

ZnS + 2HCl -> ZnCl + H2S
50mL of 0.01 M HCL added to 0.11g ZnS

A) Which reagent is in excess and by how much
B) What volume of H2S is produced at 298K in this experiment

A bottle of vinegar was labelled 4% w/v (4g per 100mL)
If 0.118 NaOH was titrated with 5mL of the vinegar, calculate the volume of NaOH used

thx in advance

fullonoob · Mar 21, 2010

adomad · Mar 21, 2010

ZnS + 2HCl -> ZnCl + H2S
50mL of 0.01 M HCL added to 0.11g ZnS

A) Which reagent is in excess and by how much
B) What volume of H2S is produced at 298K in this experiment

Find the moles of HCl in 50 mLs of 0.01M HCl and find the moles of ZnS in 0.11 grams... then subtract them to get the excess....
(Part A)
$c=\frac{n}{v}$
$\therefore n=0.01 X 50 x 10^{-3} = 0.5x10^{-3} moles_{HCl}$
$moles = \frac{mass}{Molar Mass}$
$\therefore moles_{ZnS} = \frac{0.11}{97.44}=1.1289x10^{-3}moles$
$$from the balanced equation, mole ratio between HCl: ZnS= 2:1$
$$Hence, only 0.2510^{-3}$ $moles of ZnS has reacted$
$\therefore $ ZnS is in excess by 0.8789x10^{-3}$

(Part B)
$$ mole ration of ZnS : H_2S = 1:1$
$\therefore $moles of H_2S $ $ produced is $ $moles$
$$ at 298K, any mole of gas has a volume of 24.79L$
$\therefore V_{H_2S}=24.79 x 0.2510^{-3}=0.00623$ L

adomad · Mar 21, 2010

A bottle of vinegar was labelled 4% w/v (4g per 100mL)
If 0.118 NaOH was titrated with 5mL of the vinegar, calculate the volume of NaOH used

$$ find the concentration of the vinegar.$
$c=\frac{n}{v}$
$C=$ $\frac{\frac{4}{60.032}}{0.1}$

$\therefore C_{CH_3COOH}=0.6663 M$
$$ now to find the moles of$ CH_3 COOH $ in 5 mLs of the 0.6663 M $ CH_3 COOH$
$$using $ C=\frac{n}{v}$
$n= 0.6663 X 5x10^{-3}=3.3315x10^{-3} $Moles of $ CH_3 COOH$
$$write a balanced Equation$
$NaOH+ CH_3COOH \to CH_3COONa +H_2 O (Balanced)$
$\therefore n_{NaoH}=n_{CH_3COOH}$
$\therefore n_{NaoH}= 3.3315x10^{-3}$
$C=\frac{n}{v}$
$0.118=\frac{3.3315x10^{-3}}{v}$
$\therefore V_{NaOH}= 28.2331x10^{-3}$

fullonoob · Mar 21, 2010

first step for concentration where did you get 0.1
i did 40g per 1L
n[CH3COOH] = 40/60 = 0.66

adomad · Mar 21, 2010

fullonoob said:
first step for concentration where did you get 0.1
i did 40g per 1L
n[CH3COOH] = 40/60 = 0.66

fullonoob said:
A bottle of vinegar was labelled 4% w/v (4g per 100mL)
If 0.118 NaOH was titrated with 5mL of the vinegar, calculate the volume of NaOH used

thx in advance

from there

chem calc (1 Viewer)

fullonoob

fail engrish? unpossible!

fullonoob

fail engrish? unpossible!

adomad

HSC!!

adomad

HSC!!

fullonoob

fail engrish? unpossible!

adomad

HSC!!

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